Question

Water, with a density of rho = 1125 kg/m 3 rho=1125 kg/m3 , flows in a horizontal pipe. In one segment of the pipe, the flow speed is v 1 = 6.53 m/s v1=6.53 m/s . In a second segment, the flow speed is v 2 = 1.17 m/s v2=1.17 m/s . What is the difference between the pressure in the second segment ( P 2 P2 ) and the pressure in the first segment ( P 1 P1 )?

Answer 1

The difference between the pressure in the second segment (P2) and the pressure in the first segment (P1) is 23262.75N/m^2

Step-by-step explanation:

Density (rho) of water = 1125kg/m^3

Flow speed (v1) in the first segment = 6.53m/s

From the equations of motion

v^2 = u^2 + 2gh (v = 6.53m/s, u = 0m/s, g = 9.8m/s^2)

6.53^2 = 2×9.8×h

h = 42.6409/19.6 = 2.18m

P1 = h×rho×g = 2.18×1125×9.8 = 24034.5N/m^2

Flow speed (v2) in the second segment = 1.17m/s

v^2 = 2gh

h = v^2/2gh = 1.17^2/2×9.8 = 1.3689/19.6 = 0.070m

P2 = h×rho×g = 0.070×1125×9.8 = 771.75N/m^2

P2 - P1 = 24034.5N/m^2 - 771.75N/m^2 = 23262.75N/m^2